# Instiki Sandbox

## Markdown+itex2MML Sandbox

Play around below. Your changes will, periodically, be rolled back.

### Some examples

(1)$\mathop{min} w_h p_h + w_r p_r + w_l p_l$

Here’s an equation

(2)${\int_{-\infty}^\infty e^{-a x^2/2} \mathrm{d}x} = \sqrt{\frac{2\pi}{a}}$

which we can later refer1 back to as (2).

Aligned equations:

(3)\begin{aligned} a+b &= b+a \\ a+(b+c) &= (a+b)+c \end{aligned}

The Dirac equation (boxed):

$\boxed{(i\slash{D}+m)\psi = 0}$

Here’s the table of Clifford2 algebras over $\mathbb{R}$:

$j$$0$$1$$2$$3$$4$$5$$6$$7$$8$
$\mathcal{C}\ell_{j}^-$$\mathbb{R}$$\mathbb{C}$$\mathbb{H}$$\mathbb{H}\oplus\mathbb{H}$$\mathbb{H}(2)$$\mathbb{C}(4)$$\mathbb{R}(8)$$\mathbb{R}(8)\oplus\mathbb{R}(8)$$\mathbb{R}(16)$
$\mathcal{C}\ell_{j}^+$$\mathbb{R}$$\mathbb{R}\oplus\mathbb{R}$$\mathbb{R}(2)$$\mathbb{C}(2)$$\mathbb{H}(2)$$\mathbb{H}(2)\oplus\mathbb{H}(2)$$\mathbb{H}(4)$$\mathbb{C}(8)$$\mathbb{R}(16)$

where the generators of $\mathcal{C}\ell_{j}^\pm$ satisfy

$\gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}$

and $\mathcal{C}\ell_{n+8}^\pm = \mathcal{C}\ell_n^\pm \otimes \mathbb{R}(16)$.

(4)$\lim_{n \to \infty-1} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}$

### More Examples

(5)$\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}$
(6)$\oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \mu_0 I_\text{enc}$

$H^1(\mathcal{Z}, \mathcal{O}(-k))$ Let $G=(V,E)$ be a graph, with $w:V\to [0,1]$ a weight function.

(7)$\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.$

### Theorems

###### Definition

Let $H$ be a subgroup of a group $G$. A left coset of $H$ in $G$ is a subset of $G$ that is of the form $x H$, where $x \in G$ and $x H = \{ x h : h \in H \}$.

Similarly a right coset of $H$ in $G$ is a subset of $G$ that is of the form $H x$, where $H x = \{ h x : h \in H \}$.

###### Lemma

Let $H$ be a subgroup of a group $G$, and let $x$ and $y$ be elements of $G$. Suppose that $x H \cap y H$ is non-empty. Then $x H = y H$.

###### Proof

Let $z$ be some element of $x H \cap y H$. Then $z = x a$ for some $a \in H$, and $z = y b$ for some $b \in H$. If $h$ is any element of $H$ then $a h \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup of $G$. But $z h = x(a h)$ and $xh = z(a^{-1}h)$ for all $h \in H$. Therefore $z H \subset x H$ and $x H \subset z H$, and thus $x H = z H$. Similarly $y H = z H$, and thus $x H = y H$, as required.

###### Lemma

Let $H$ be a finite subgroup of a group $G$. Then each left coset of $H$ in $G$ has the same number of elements as $H$.

###### Proof

Let $H = \{ h_1, h_2,\ldots, h_m\}$, where $h_1, h_2,\ldots, h_m$ are distinct, and let $x$ be an element of $G$. Then the left coset $x H$ consists of the elements $x h_j$ for $j = 1,2,\ldots,m$. Suppose that $j$ and $k$ are integers between $1$ and $m$ for which $x h_j = x h_k$. Then $h_j = x^{-1} (x h_j) = x^{-1} (x h_k) = h_k$, and thus $j = k$, since $h_1, h_2,\ldots, h_m$ are distinct. It follows that the elements $x h_1, x h_2,\ldots, x h_m$ are distinct. We conclude that the subgroup $H$ and the left coset $x H$ both have $m$ elements, as required.

###### Theorem

(Lagrange’s Theorem). Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$.

###### Proof

Each element $x$ of $G$ belongs to at least one left coset of $H$ in $G$ (namely the coset $x H$), and no element can belong to two distinct left cosets of $H$ in $G$ (see Lemma 1). Therefore every element of $G$ belongs to exactly one left coset of $H$. Moreover each left coset of $H$ contains $|H|$ elements (Lemma 2). Therefore $|G| = n |H|$, where $n$ is the number of left cosets of $H$ in $G$. The result follows.

###### Corollary

Let $x$ be an element of a finite group $G$. Then the order of $x$ divides the order of $G$.

###### Theorem

Let $f : \Delta \longrightarrow \Delta,$ where $\Delta=\{z\in\mathbb{C}: \vert z \vert \lt 1\}$, be analytic with $a \in \Delta$. Then

$\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert$

for all $\vert z \vert \le 1$ and

$\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.$

Furthermore, equality holds iff $f$ realizes a conformal mapping of $\Delta$ onto itself.

###### Proof

Let $w=\frac{z-a}{1-\overline{a}z}$ and put $\phi(w)=\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}$. Define for $\abs{b}\lt 1$ $C_b(z)=\frac{z-b}{1-\overline{b}z}.$ All conformal maps from $\Delta$ to itself, sending $b$ to $0$, are of the form $C_b(z)e^{i\gamma}$ for $\gamma\in[0,2\pi].$ In this notation, $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w),$ where $C_a^{-1}$ is the inverse of $C_a$ as a function. Note that $C_a(z)$ is conformal, so it has an inverse. It is clear that $\phi(0)=C_{f(a)}\circ f \circ C_a^{-1}(0)=C_{f(a)}(f(a))=0$. Since $C_a^{-1}: \Delta \longrightarrow \Delta$ and $f : \Delta \longrightarrow \Delta$ and $C_{f(a)}: \Delta \longrightarrow \Delta,$ then $\vert\phi(w)\vert\lt 1$ for $\vert w\vert \lt 1 .$ Applying Schwarz’s lemma, we obtain $\vert\phi(w)\vert\le \vert w \vert$ for $\vert w \vert \le 1$. Furthermore, if equality holds, then $f(z)=e^{i\gamma'} z$ for $\gamma'\in [0,2\pi]$. Therefore,

(8)$\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert$

for all $\vert z \vert\le 1.$ Rearranging, we obtain

$\left\vert\frac{f(z)-f(a)}{z-a}\right\vert\le\left\vert\frac{1-\overline{f(a)}f(z)}{1-\overline{a}z}\right\vert.$

If we take the limit as $z$ tends to $a$, we obtain

$\left\vert f'(a)\right\vert \le \left\vert\frac{1-\vert f(a)\vert ^2}{1-\vert a \vert^2}\right\vert=\frac{1-\vert f(a)\vert^2}{1-\vert a \vert^2},$

or

$\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.$

As said above, if equality holds in (8), then Schwarz’s lemma tells us that $\phi(w)=e^{i\gamma'}w$. Thus, $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma'}w,$ so $f(z)=C_{f(a)}^{-1}(e^{i\gamma'}C_a(z))$. Since $e^{i\gamma'}C_a(z)$ is conformal, $C_{f(a)}^{-1},$ the inverse function of $C_{f(a)}$, is conformal, and a composition of conformal maps is conformal, then $f$ is a conformal map of $\Delta$ onto itself. Conversely, if $f$ is a conformal map of $\Delta$ onto itself, then $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma}C_b(w),$ since a composition of conformal maps is conformal and because all conformal maps from $\Delta$ onto itself are of the form $e^{i\gamma}C_b(w).$ We also know that $\phi(0)=0,$ so $b=0$. Therefore,

$\phi(w)=e^{i\gamma}C_0(w)=e^{i\gamma}w \Leftrightarrow \vert\phi(w)\vert=\vert w \vert \Leftrightarrow \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert=\left\vert\frac{z-a}{1-\overline{a}z}\right\vert$

for all $\vert z \vert\le 1$. In sum, equality holds in (8) iff $f$ is a conformal map from $\Delta$ to itself.

###### Remark

Someone needs to code \abs, i.e. \left\vert # \right\vert. This is terribly annoying. Also, when using , why must one enter a line before and after.

#### SVG graphics, created in SVG-Edit (Instiki 0.19 feature):

$E_\Sigma$

$K^0\overline{K}^0$ Mixing

#### Yet More examples

• SVG:
• Animated SVG
• Complicated commutative diagrams (equations in SVG)
• SVG in equations.

In $SU(3)$, $\begin{svg} \end{svg}\includegraphics[width=2em]{young1} \otimes \begin{svg} \end{svg}\includegraphics[width=1em]{young2} = \begin{svg} \end{svg}\includegraphics[width=2em]{young3} \oplus \begin{svg} \end{svg}\includegraphics[width=3em]{young4}$.

• Cases:
$r_{a+1} = \begin{cases} 0 & \text{with prob.}\quad \exp(-\theta r_a) \\ \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a) \\ \end{cases}$
• Stretchy Brackets:
$q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}$
$\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i } = \alpha \sum_{i = 1}^N {f(x_i )w_i } + \beta \sum_{i = 1}^N {g(x_i )w_i }$
• Linearity of Integrals
${\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx} + \beta \int_a^b {g(x)\,dx} }$
• Can we talk about $x_i^2$ inline? What about $\int_a^b x^2\,dx$? Inline fractions $\frac{x-x_2}{x_1-x_2}$?

• Big fractions

$p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}$
• Diagram
$\begin{matrix} P_1(Y) &\to& P_1(X) \\ \downarrow &\Downarrow^\mathrlap{\sim}& \downarrow \\ T' &\to& T \end{matrix}$

(9)\array{\arrayopts{\align{center}} \begin{svg} \end{svg} } \equiv \left({U(k)}^{n_1},\{v_i\}\right)

“This is my text”, says Anymouse.

Ruby code example:

    class Person
def initialize(name, age)
@name, @age = name, age
end
def <=>(person) # Comparison operator for sorting
@age <=> person.age
end
def to_s
"#@name (#@age)"
end
end

group = [
Person.new("Bob", 33),
Person.new("Chris", 16),
Person.new("Ash", 23)
]

puts group.sort.reverse

char code test:

• 日本語のテスト
• きちんと処理できることを確認中

これで どうだ !

1. You can also refer to it as (2). Chacun à son goût!.